A camper dives from the edge of a swimming pool at water level with a speed of 8.0 m/s at an angle of 30.0° above the horizontal. (a) How long is the diver in the air? (b) How high is the diver in the air? (c)How far out in the pool does the diver land?

Answer:

[tex]r= 2.03*10^7m[/tex]

V = 1.45x10^3 m/s

Explanation:

number of days in sec = 1.02days * 86400s = 88128 s              

Mass of Mars is 6.42*10^23 Kg

gravitational constant G = 6.674*10^{-11}

[tex]T^{2} = \frac {4pi^2}{GM} *r^3[/tex]

[tex]\frac {88128^2}{(9.22*10^{-13})} = r^3[/tex]

[tex]r= 2.03*10^7m[/tex]

(b) [tex]V=\frac{2 \pi*r}{T}[/tex]

[tex]V=\frac{(2 \pi(2.03*10^7))}{(88128)} = 1.45x10^3 m/s[/tex]

V

Answer:

The pressure that the skater exerts on ice if they are balanced on a single skate is P= 18,312 * 10⁶ Pa = 180.72 atm.

Explanation:

width= 0.025cm = 2.5 * 10⁻⁴ m

lenght= 15 cm = 0.15m

m= 70 kg

g= 9.81 m/s²

S= width * lenght

S= 3.75 * 10⁻⁵ m²

F= m*g

F= 686.7 N

P= F/S

P= 18.312 * 10⁶ Pa= 180.72 atm

Answer:

95.47 C

Explanation:

Heat added = mass × heat capacity × rise in temperature

So, rise in temperature =

20000 / (0.05 × 4190)

Rise in temperature = 95.47 C

To calculate the final temperature of [tex]50 g[/tex]  of water heated with [tex]20000 J[/tex], you use the equation [tex]Q = mc\Delta T[/tex]. The calculated temperature change is [tex]95.5^{\circ}C[/tex], and adding this to an initial temperature of [tex]25^{\circ}C[/tex] gives [tex]120.5^{\circ}C[/tex]. Because water boils at [tex]100^{\circ}C[/tex], the final temperature would be limited to [tex]100^{\circ}C[/tex].

To calculate the final temperature of [tex]50 g[/tex] of water heated with [tex]20000 J[/tex] of energy, we can use the equation:

[tex]Q = mc\Delta T[/tex]

where:

Heat energy, [tex]Q = (20000 J)[/tex]

Mass of water. [tex]m = (50 g = 0.05 kg)[/tex]

Specific heat capacity of water, [tex]c = (4190 \, \text{J/kg°C})[/tex]

Change in temperature, [tex]\Delta T = (T_{\text{final}} - T_{\text{initial}})[/tex]

Rearrange the equation to solve for[tex]\Delta T[/tex]:

[tex]\Delta T = \frac{Q}{mc}[/tex]

Substitute the values:

[tex]\Delta T = \frac{20000 \, \text{J}}{0.05 \, \text{kg} \times 4190 \, \text{J/kgC}}[/tex]

[tex]\approx 95.5^{\circ}C[/tex]

If the initial temperature of the water was[tex]25^{\circ}C[/tex] (for example), then the final temperature would be:

[tex]T_{\text{final}} = T_{\text{initial}} + \Delta T\\T_{\text{final}} = 25^\circ \text{C} + 95.5^\circ \text{C} = 120.5^\circ \text{C}[/tex]

However, remember that water boils at 100°C under standard pressure, so the actual final temperature would be limited to .

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar,  F = 534 N

Co-efficient of kinetic friction, [tex]\mu_k[/tex] = 0.3

Now,

The force against the kinetic friction is given as:

[tex]f = \mu_k N = u_k Mg[/tex]

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

[tex]f = 0.3\times108\times9.8[/tex]

or

[tex]f = 317.52N[/tex]

Now, the net force on to the metal safe is

[tex]F_{Net}= F-f[/tex]

Substituting the values in the equation we get

 [tex]F_{Net}= 534N-317.52N[/tex]

or

[tex]F_{Net}= 216.48[/tex]

also,

[tex]F_{Net}= M\times [/tex]acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=[tex] \frac{F_{Net}}{M} [/tex]

or

 acceleration of the safe=[tex] \frac{216.48}{108} [/tex]

or

acceleration of the safe=[tex] 2.00 m/s^2 [/tex]

Hence, the acceleration of the metal safe will be  2.00 m/s²

The acceleration of the safe is determined by factoring in the force exerted by the burglar, the static friction that initiates movement, and the kinetic friction that must be overcome when the safe is in motion. When these factors are calculated, the acceleration comes out to be approximately 2 m/s².

The subject in question deals with two types of force: the force applied by the burglar and the frictional force which acts against the direction of the motion. The gravitational force acting on the safe, also known as its weight, can be calculated by multiplying the safe's mass (108 kg) with the acceleration due to gravity (approx. 9.80 m/s²), which gives us a value of 1058.4 N. This weight also represents the normal force, as the safe is on a horizontal plane.

The maximum force of static friction, calculated using the formula ƒs_max = μsN (where μs is the coefficient of static friction and N is the normal force), turns out to be 0.4 * 1058.4 N = 423.36 N. This implies the burglar needs to exert a force greater than this to overcome the static friction and set the safe in motion.

Given that the burglar can apply a maximum force of 534 N, this is significantly greater than the static friction, inducing motion in the safe. Once the safe is moving, it's the force of kinetic friction that matters. Calculating this force gives us 0.3 * 1058.4 N = 317.52 N. This is the force that has to be overcome to maintain the safe in motion.

Using Newton's second law (F = ma), we can determine the acceleration by subtracting kinetic friction from the applied force and dividing it by the mass of the safe. This gives us an acceleration of (534N - 317.52N) / 108kg = 2 m/s². Therefore, the safe would indeed move, and its acceleration would be 2 m/s².

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Answer:

1+5x+10x^2+10x^3+5x^4+x^5

Explanation:

You just make it (1+x)(1+x)(1+x)(1+x)(1+x) and multiply it out until it's all one big term.

Answer:

T=78.48 N

Explanation:

We know that the moment developed at a hinge equals zero

Thus summing moments about hinge we have

(See attached figure)

[tex]2.0\times g\times \frac{1.5}{2}+1.50\times 5\times g-T\times \frac{3}{4}1.5=0\\\\Solving\\\\T=\frac{1}{1.125}(88.29)\\\\T=78.48N[/tex]

Answer:

Explanation:

To estimate the focal length of a convex lens follow the following steps.

1. take a convex lens.

2. Stand near a window which is just opposite to a wall.

3. Look at a tree which is far away from the window by the convex lens.

4. focus the image of the tree on the wall which is opposite to the window.

5. You wll observe that by changing the position of convex lens a sharp and inverted and small image is seen on the wall.

5. Now measure the distance between the lens and the wall.

7. This distance is the rough focal length of the convex lens.

To estimate the focal length of an unmarked convex lens, you can hold the lens in front of a bright object and project the image onto a blank wall until it's clear and at its smallest size. The distance between the lens and wall is the focal length. Refraction of light in materials like water changes their lens properties.

To estimate the focal length of an unmarked convex lens quickly, you can employ a simple method using readily available materials. Here's a step-by-step explanation of this experimental process:

This method is based on the fact that when an object is placed at a great distance from a converging lens (much greater than the focal length), the image is formed at the focal point on the other side of the lens. By finding this point of clear image formation, you effectively measure the focal length of the convex lens.

Impact of Refraction

When you fill a glass or plastic bottle with water, it can act as a converging lens due to the refraction of light. The water inside the bottle has a different index of refraction compared to the air, which allows the bottle to focus light and form images like a lens. The curvature of the bottle and the water's index of refraction contribute to the lens properties and focal length of the water bottle lens.

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

        V ∝ T

That is

       [tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

Here given that

      V₁ = 0.20 cubic meter

      T₁ = 333 K

      T₂ = 533 K

Substituting

      [tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3[/tex]

New volume of the gas  = 0.3198 m³

Option C is the correct answer.

Answers:

(a) [tex]0.0073kg[/tex]

(b) Volume gold: [tex]3.79(10)^{-7}m^{3}[/tex], Volume cupper: [tex]7.6(10)^{-8}m^{3}[/tex]

(c) [tex]17633.554kg/m^{3}[/tex]

Explanation:

We are told the total mass [tex]M[/tex] of the coin, which is an alloy  of gold and copper is:

[tex]M=m_{gold}+m_{copper}=7.988g=0.007988kg[/tex]   (1)

Where  [tex]m_{gold}[/tex] is the mass of gold and [tex]m_{copper}[/tex] is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats [tex]K[/tex] and mass is:

[tex]K=24\frac{m_{gold}}{M}[/tex]   (2)

Finding [tex]{m_{gold}[/tex]:

[tex]m_{gold}=\frac{22}{24}M[/tex]   (3)

[tex]m_{gold}=\frac{22}{24}(0.007988kg)[/tex]   (4)

[tex]m_{gold}=0.0073kg[/tex]   (5)  This is the mass of gold in the coin

The density [tex]\rho[/tex] of an object is given by:

[tex]\rho=\frac{mass}{volume}[/tex]

If we want to find the volume, this expression changes to: [tex]volume=\frac{mass}{\rho}[/tex]

For gold, its volume [tex]V_{gold}[/tex] will be a relation between its mass [tex]m_{gold}[/tex]  (found in (5)) and its density [tex]\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}[/tex]:

[tex]V_{gold}=\frac{m_{gold}}{\rho_{gold}}[/tex]   (6)

[tex]V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}[/tex]   (7)

[tex]V_{gold}=3.79(10)^{-7}m^{3}[/tex]   (8)  Volume of gold in the coin

For copper, its volume [tex]V_{copper}[/tex] will be a relation between its mass [tex]m_{copper}[/tex]  and its density [tex]\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}[/tex]:

[tex]V_{copper}=\frac{m_{copper}}{\rho_{copper}}[/tex]   (9)

The mass of copper can be found by isolating [tex]m_{copper}[/tex] from (1):

[tex]M=m_{gold}+m_{copper}[/tex]  

[tex]m_{copper}=M-m_{gold}[/tex]  (10)

Knowing the mass of gold found in (5):

[tex]m_{copper}=0.007988kg-0.0073kg=0.000688kg[/tex]  (11)

Now we can find the volume of copper:

[tex]V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}[/tex]   (12)

[tex]V_{copper}=7.6(10)^{-8}m^{3}[/tex]   (13)  Volume of copper in the coin

Remembering density is a relation between mass and volume, in the case of the coin the density [tex]\rho_{coin[/tex] will be a relation between its total mass [tex]M[/tex] and its total volume [tex]V[/tex]:

[tex]\rho_{coin}=\frac{M}{V}[/tex] (14)

Knowing the total volume of the coin is:

[tex]V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}[/tex] (15)

[tex]\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}[/tex] (16)

Finally:

[tex]\rho_{coin}=17633.554kg/m^{3}}[/tex] (17)  This is the total density of the British sovereign coin

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

[tex]\Delta P.E=mg(h_{2}-h_{1})[/tex]

[tex]\Delta P.E=10000\times9.8\times(10000-0)[/tex]

[tex]\Delta P.E=10000\times9.8\times10000[/tex]

[tex]\Delta P.E=980000000\ J[/tex]

[tex]\Delta P.E=980\ MJ[/tex]

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

[tex]\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)[/tex]

[tex]\Delta K.E=148298642\ J[/tex]

[tex]\Delta K.E=148.3\ MJ[/tex]

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

In physics, inertia refers to an object's resistance to a change in motion, and it is directly proportional to the object's mass. This means the object with the greater mass will have more inertia. Given the provided options, the 7-kilogram object at rest has the greatest inertia because it has the most mass.

The subject in question relates to inertia, a concept in physics that describes an object's resistance to a change in motion. Inertia is directly proportional to an object's mass, meaning an object with more mass exhibits greater inertia. Hence, considering the options a. A 2-kilogram object moving at 5 m/s, b. A 5-kilogram object moving at 3 m/s, c. A 7-kilogram object at rest, and d. A 3-kilogram object moving at 4 m/s, the object that has the greatest inertia would be c. A 7-kilogram object at rest. This is because it has the greatest mass out of all the options.

Inertia is associated with Newton's first law of motion, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. This can be seen in daily life - for example, it's more difficult to push a heavy truck into motion than a small toy because the truck has a greater mass and hence more inertia.

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Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

Using the given intensity of the electromagnetic wave and the fundamental constants, we substitute into the formula B = √(2I/μoc²). The resulting amplitude of the magnetic field is approximately 7.98 × 10-⁶ Tesla.

The student is asking for the calculation of the amplitude of a magnetic field given the intensity of an electromagnetic wave. This belongs to the realm of Physics, specifically electromagnetism. We can use the formula I = 1/2μoc²B² to solve for this, where I is the intensity, μo is the permeability of free space, c is the speed of light, and B is the maximum strength of the magnetic field.

First, we rearrange the formula to solve for B, yielding B = √(2I/μoc²). Substituting the given values, we get B = √(2*80x10⁶ W/m²/(4π×10−7 T m/A * (3.0×10⁸ m/s)²). Calculating this gives us a magnetic field amplitude of approximately 7.98 × 10-⁶ T.

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Final answer:

To find the amplitude of the magnetic field for an electromagnetic wave with a given intensity, use the formula I = cε0B2. For an intensity of 80 MW/m2, the calculated magnetic field amplitude is approximately 1.69×10−3 T (1.69 mT).

Explanation:

The intensity I of an electromagnetic wave can be related to its magnetic field B using the relationship I = cε0B2, where c is the speed of light in a vacuum, and ε0 is the permittivity of free space. Given that the intensity I is 80 MW/m2, we can use the given values for c (3.0×108 m/s) and ε0 (8.85×10−12C2/N·m2) to find the amplitude of the magnetic field.

First, rearrange the formula to solve for B:

B = √(I / (cε0))

Substitute the given values:

B = √(80×106 W/m2 / (3.0×108 m/s × 8.85×10−12 C2/N·m2))

After performing the calculations:

B = 1.69×10−3 T

Therefore, the amplitude of the magnetic field for an electromagnetic wave with an intensity of 80 MW/m2 is approximately 1.69 mT (milliteslas).

Answer:

Charge, [tex]q=-2.14\times 10^{-5}\ C[/tex]

Explanation:

It is given that,

Mass of the charged particle, m = 1.44 g = 0.00144 kg

Electric field, E = 660 N/C

We need to find the charge of that particle to remain stationary when placed in a downward-directed in the given electric field such that its weight is balanced by the electrostatic force i.e.

[tex]mg=qE[/tex]

[tex]q=\dfrac{mg}{E}[/tex]

[tex]q=\dfrac{0.00144\ kg\times 9.81\ m/s^2}{660\ N/C}[/tex]

q = 0.0000214 C

[tex]q=2.14\times 10^{-5}\ C[/tex]

Since, the electric field is acting in downward direction, so the electric force will act in opposite direction such that they are in balanced position. Hence, the charge must be negative.

i.e. [tex]q=-2.14\times 10^{-5}\ C[/tex]

Answer:

Option (D)

Explanation:

The definition of specific heat is given below

The amount of heat required to raise the temperature of 1 kg substance by 1 degree Celsius.

Q = m c (T2 - T1)

c = Q / m × (T2 - T2)

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

Answer:

3.26 x 10^9 A

Explanation:

I1 = 7.2 A, r = 18.1 cm = 0.181 m, F/l = 2.6 x 10^4 N/m

Let teh current in other wire is I2.

Use the formula of force per unit length

[tex]F / l = \frac{\mu _{0}}{4\pi }\times \frac{2 I_{1}I_{2}}{r}[/tex]

[tex]2.6 \times 10^{4} = 10^{-7}\frac{2 \times 7.2I_{2}}{0.181}[/tex]

I2 = 3.26 x 10^9 A

Final answer:

The current in the second wire is 2.0 A.

Explanation:

The current in the second wire is 2.0 A.

Given that the force per unit length between the wires is directly proportional to the product of their currents, we can set up a proportion to find the current in the second wire:

[tex](7.2 A) / (2.6 x 10^4 N/m) = (x A) / (2.6 x 10^4 N/m)[/tex]

Solving for x gives x = 2.0 A, which is the current in the second wire.

Answer:

The wavelength of light is [tex]4.53\times10^{-7}\ m[/tex]

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

[tex]d \sin\theta=n\lambda[/tex].....(I)

The separation of the slits

[tex]d = \dfrac{1}{N}[/tex]

[tex]d=\dfrac{1}{5000}[/tex]

[tex]d=2\times10^{-6}\ m[/tex]

Now put the value in equation (I)

[tex]2\times10^{-6}\sin13.1^{\circ}=\lambda[/tex]

Here, n = 1

[tex]\lambda=4.53\times10^{-7}\ m[/tex]

Hence, The wavelength of light is [tex]4.53\times10^{-7}\ m[/tex]

Answer:

1.586m

Explanation:

let 'u' be the velocity of the roof tile when it reaches the window

now using the equation of motion

[tex]s=ut+\frac{1}{2}at^{2}[/tex]

where s= distance travelled

u=velocity

a=acceleration of the object

t=time taken to travel the distance 's'

given:

s=1.7m (distance covered to pass the window)

t=0.25s (Time taken to pass the window)

a=g=9.8m/s^2 (since the roof tile is moving under the action of gravity)

thus, substituting the values in the above equation we get

[tex]1.7=u\times 0.25+\frac{1}{2}\times 9.8\times0.25^{2}[/tex]

u=5.575m/s

This is the velocity when the tile  touches the window top.

Let's take this in second scenario as the tile's final velocity(v).

Now we have another equation of motion as

[tex]v^{2}-u^{2}=2as[/tex]

initial speed (when starts to fall) will be zero.

So the distance travelled (h) i.e the height from which the tile falls from the top of the window is given by,

substituting the values in the above equation, we get

[tex]5.575^{2}-0^{2}=2\times 9.8\times h[/tex]

[tex]h=\frac{5.575^{2}-0^{2}}{2\times 9.8}[/tex]

h=1.586m

Hence, the window roof is 1.586m far away from the roof

Final answer:

To determine the height above the window where the roof tile fell, we use the kinematic equations with the given time and the height of the window. After calculating the velocity at the bottom of the window and the time taken to pass, we find the total distance fallen from the roof and subtract the window height to get the height above the window.

To determine how far above the top of the window the roof is where the roof tile fell, we will apply kinematic equations for uniformly accelerated motion, which in this case, is due to gravity. The tile falls past the window in 0.25 s, covering a distance of 1.7 m.

First, we calculate the velocity of the tile at the bottom of the window using the equation v = v_0 + at, where v_0 is the initial velocity (0 m/s since it falls from rest), a is the acceleration due to gravity (9.81 m/s2), and t is the time it takes to pass the window. We assume the tile's speed at the top of the window is approximately the same as at the bottom since the window height is relatively small. This gives us a calculation to find the speed at the bottom: v = 0 m/s + (9.81 m/s2)(0.25 s). We'll use this speed as an average speed to simplify the calculation.

Using d = vt, where d is the distance covered (1.7 m) and v is the average velocity, we solve for the time it takes to pass the window. This provides half the time of passage, thus t = d/v. Then we use the time to find the total distance fallen from the roof, d_total, using the equation d_total = [tex]0.5at^2[/tex].  Finally, we subtract the window height from d_total to find the height above the window where the tile fell, which is the answer to the question.

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

  m=8/32=0.25

To find the spring constant

Kx=mg    (given that c=6 in and mg=8 lb)

K(0.5)=8               (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F  

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved. The steady-state response of this system will be y= 240/901 cos 2t+ 8/901 sin 2t.

A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved following an external stimulus.

The given data in the problem will bge;

C=0.25 lb.s/ft

Weight is defined as the product of mass and gravity.

[tex]\rm{m=\frac{W}{g} }\\\\\rm{m=\frac{8}{32}[/tex]

[tex]\rm m=0.25[/tex]

Spring constant is defined as the ratio of force per unit displaced length.

The spring force is balanced by the weight;

[tex]\rm Kx=mg\\\\ \rm x= \frac{mg}{K} \\\\ \rm x=\frac{8}{0.5} lb/ft[/tex]

The equation for the spring-mass system is given by;

[tex]\rm {my''+Cy'+Ky=F }[/tex]

[tex]\rm 0.25 y"+0.25 y'+16 y=4 cos 20 t[/tex]

Steady-state equation;

[tex]\rm y=A cos 2t+ B sin 2t[/tex]

For finding the value of A and B

[tex]\rm y'= -2A sin 2t+2B cos 2ty"=-4A cos 2t-4B sin 2t[/tex]

By putting the value we got

[tex]\rm 0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t[/tex]

The value of cofficient obtained from the equation

[tex]30 A+ B=8[/tex]

Getting the value as

[tex]A= \frac{240/901}\\\\ B=\frac{8}{901}[/tex]

The steady-state response got

[tex]y= 240/901 cos 2t+ 8/901 sin 2t[/tex]

Hence the steady-state response of this system.y= 240/901 cos 2t+ 8/901 sin 2t

To learn about the steady-state response refer to the link;

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The magnetic force acting on a charged particle moving perpendicular to the field is:

[tex]F_{b}[/tex] = qvB

[tex]F_{b}[/tex] is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

[tex]F_{c}[/tex] = mv²/r

[tex]F_{c}[/tex] is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set [tex]F_{b}[/tex] equal to [tex]F_{c}[/tex] and solve for v:

qvB = mv²/r

v = qBr/m

Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:

W = KE

W is work, KE is kinetic energy

W = Vq

KE = 0.5mv²

Therefore:

Vq = 0.5mv²

Substitute v = qBr/m and solve for V:

V = 0.5qB²r²/m

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

B = 0.750T

q = 1.60×10⁻¹⁹C (proton charge)

r = 1.84×10⁻²m

Plug in the values and solve for V:

V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷

V = 9120V

The proton, initially accelerated through a potential difference V and then making a circular path in a magnetic field, allows us to calculate that V is approximately 8.74 x 10^5 volts.

The question involves the concept of a proton moving in a magnetic field after being accelerated through a voltage V. As a proton enters a magnetic field perpendicular to its path, it follows a circular arc. Let's use the known concepts of physics to derive the required voltage.

The radius of the proton's path can be calculated using the Lorentz force formula: F = qvB = mv^2/r, where q is the charge of the proton, v is the velocity, B is the magnetic field, m is the proton's mass, and r is the radius. From this equation, we can express velocity as v = qBr/m.

Next, we know that the kinetic energy of the proton (K.E.) equals the work done on it, which is the voltage times the charge of the proton: K.E. = qV. Also, the kinetic energy can be expressed as K.E. = 1/2 mv^2.

Equating these two forms of kinetic energy, we get 1/2 mv^2 = qV. Substituting our expression for velocity from above, we find V = ( q^2 B^2 r^2) / (2 m).

Plug in the known values: q = 1.60 x 10^-19 C, B = 0.750 T, r = 1.84 x 10^{-2} m (converted from cm to m), and m = 1.67 x 10^-27 kg (mass of a proton), we find that V is approximately 8.74 x 10^5 volts.

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Answer:

The angular position of the wheel after 2.7 seconds is θf= 11.35 rad.

Explanation:

θi= 8.8 rad

ωi= 0 rad/seg

α= 0.7 rad/seg²

θf= θi + ωi * t + α * t² / 2

θf= 11.35 rad

Answer:

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-0.44m/s[/tex]

Explanation:

Displacement,

          [tex]s(t)=\frac{6}{t^2}[/tex]

Velocity is given by

          [tex]v(t)=\frac{ds}{dt}=\frac{d}{dt}\left ( \frac{6}{t^2}\right )=-\frac{12}{t^3}[/tex]

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-\frac{12}{1^3}=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-\frac{12}{2^3}=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-\frac{12}{3^3}=-0.44m/s[/tex]

The velocity of the particle for different times (t = a, 1, 2, and 3 seconds) is found by differentiating the displacement equation s = 6/t^2 and applying the values of t to get the velocities: v(a) = -12/a^3, v(1) = -12 m/s, v(2) = -1.5 m/s, and v(3) ≈ -0.44 m/s.

The question you've asked is about finding the velocity of a particle moving in a straight line where its displacement is given by s = 6/t2, and t is the time in seconds. To find the velocity (v) at any given time, we need to take the derivative of the displacement with respect to time. So the derivative of s with respect to t gives us v = -12/t3. Let's apply this formula for t = a, 1, 2, and 3 seconds.

Note that the negative sign indicates the direction of velocity is opposite to the direction assumed as positive in the displacement equation.

Answer:

[tex]v = 567.2 km/h[/tex]

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

[tex]distance = 2\pi r[/tex]

[tex]distance = \pi D[/tex]

now we will have the time to complete the rotation given as

[tex]t = 6 days[/tex]

[tex]t = 6 (24 h) = 144 h[/tex]

now the speed is given by

[tex]speed = \frac{distance}{time}[/tex]

[tex]speed = \frac{\pi D}{t}[/tex]

[tex]speed = \frac{\pi(26000 km)}{144}[/tex]

[tex]v = 567.2 km/h[/tex]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

Answer:

Explanation:

Givens

Time taken to go down + time taken for the sound to come up = 1.17 seconds.

d = 7.92 m

a = 9.81 m/s^2

t (see below)

vi = ???

Solution to How long it takes to come back up.

v = 343 m/s

d = 7.92 meters

t = ?

t = d/v

t = 7.92 m / 343 m/s

t = 0.0231 seconds.

Solution to time taken to go down.

Time_down = 1.17 - 0.0231

time_down = 1.147 seconds

Solution to vi

d = vi*t + 1/2 a t^2

7.92 = vi*1.147 + 1/2 * 9.81 * 1.147^2

7.92 = vi*1.147 + 6.452                    Subtract 6.452 from both sides.

7.92 - 6.452 = 1.147*vi

1.468 = 1.147 * vi                              Divide by 1.147

1.468 / 1.147 = vi            

1.279 m/s = vi

Answer:

14.7 Volt

Explanation:

C1 = 2.74 F, C2 = 7.46 F, C3 = 10.1 F

Here C1 and C3 are in parallel

So, Cp = C1 + C3 = 2.74 + 10.1 = 12.84 F

Now Cp and C2 are in series

1 / C = 1 / Cp + 1 / C2

1 / C = 1 / 12.84 + 1 / 7.46

C = 4.72 F

Let q be the total charge

q = C V = 4.72 x 40 = 188.8 C

Voltage across C2

V2 = q / C2 = 188.8 / 7.46 = 25.3 V

Voltage across C2 or c3

V' = V - V2 = 40 - 25.3

V' = 14.69

V' = 14.7 Volt

Answer:

The magnitude of the angular momentum of the cylinder and the rotational kinetic energy of the cylinder are 0.0205 Kgm²/s and 0.01317 J

Explanation:

Given that,

Moment of inertia = 0.016 kg m²

Radius = 6.0

Linear speed = 7.7 m/s

We need to calculate the angular momentum

Using formula of angular momentum

[tex]L=I\omega[/tex]

Where, L = angular momentum

I = moment of inertia

[tex]\omega[/tex] =angular velocity

Put the value into the formula

[tex]L=0.016\times\dfrac{7.7}{6.0}[/tex]

[tex]L=0.0205\ Kg m^2/s[/tex]

We need to calculate the rotational kinetic energy of the cylinder

Using formula of Rotational kinetic energy

[tex]K.E=\dfrac{1}{2}\times I\omega^2[/tex]

[tex]K.E= \dfrac{1}{2}\times I\times(\dfrac{v}{r})^2[/tex]

[tex]K.E= \dfrac{1}{2}\times0.016\times(\dfrac{7.7}{6.0})^2[/tex]

[tex]K.E=0.01317\ J[/tex]

Hence, The magnitude of the angular momentum of the cylinder and the rotational kinetic energy of the cylinder are 0.0205 Kg m²/s and 0.01317 J

Answer:

Axial flow pumps can usually handle large flow rates. They hence have high specific speed - b.

Axial flow pumps are capable of handling large flow rates and have a high specific speed, which is a contrast to positive displacement pumps like diaphragm pumps that deliver a constant flow regardless of pressure.

Axial flow pumps are designed to handle large flow rates, which indicates their ability to move a high volume of fluid. The specific speed of a pump describes how efficient it is at handling different flow rates. In this context, axial flow pumps that can handle large flow rates usually have low specific speed (a) as they are more effective at high flow rates than pumps with higher specific speeds.

Axial flow pumps are designed to handle very large flow rates and are characterized by their ability to deliver fluid primarily in a direction parallel to the pump shaft. This capability allows them to operate efficiently under conditions where a significant volume of liquid needs to be moved across relatively short distances with lower head (pressure) requirements. The concept of specific speed is important when discussing pump performance. It is a dimensionless number that describes a pump's shape and characteristics based on its speed, flow rate, and head. Given that axial flow pumps are optimal for applications with high flow rates and lower head, they are distinguished by a high specific speed. This is in contrast to positive displacement pumps, like diaphragm pumps, which deliver a constant flow for each revolution of the pump shaft, regardless of changes in pressure, highlighting a key difference in operation and efficiency between pump types.

Answer:

19.09 m/s

Explanation:

u = 0, h = 18.6 m

Use third equation of motion

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 18.6

v = 19.09 m/s